# How do you solve and find the value of tan(sin^-1(1/2))?

Nov 26, 2016

$\tan \left({\sin}^{- 1} \left(\frac{1}{2}\right)\right) = \frac{1}{\sqrt{3}}$

#### Explanation:

Inverse trigonometric ratios are the inverse functions of the trigonometric functions. As an example, $y = \sin x$ means sine of $\angle x$ is $y$. As an inverse trigonometric ration this means that $\arcsin y = x$ or ${\sin}^{- 1} y = x$.

As $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$, we have ${\sin}^{- 1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$

and hence

$\tan \left({\sin}^{- 1} \left(\frac{1}{2}\right)\right) = \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$